Question

The concentration of 1L of \( CaCO_3 \) solution is 1000 ppm. What is its concentration in mol \( L^{-1} \)?
(Ca = 40 u, O = 16 u, C = 12 u)

• A. \( 10^{-3} \)
• B. \( 10^{-1} \)
• C. \( 10^{-4} \)
• D. \( 10^{-2} \)

Hint:

To convert ppm to molarity, use the relation: \[ \text{Molarity} = \frac{\text{ppm value}}{\text{Molar mass} \times 1000} \]

Answer 1

The Correct Option is D

Step 1: Calculate the molar mass of CaCO₃.
We are given:
 

• Ca = 40 μ
• C = 12 μ

We know that O = 16 μ.
Molar Mass of CaCO₃ = Ca + C + 3 × O
Molar Mass of CaCO₃ = 40 + 12 + 3 × 16 = 40 + 12 + 48 = 100 g/mol

 

Step 2: Convert ppm to mg/L.
We are given that 1 L CaCO₃ concentration = 1000 ppm.
1 ppm = 1 mg/L
Therefore, 1000 ppm = 1000 mg/L

Step 3: Convert mg/L to g/L.
1000 mg/L = 1 g/L.

Step 4: Convert g/L to mol/L
We have 1 g/L of CaCO₃.
Molar mass of CaCO₃ = 100 g/mol.
Concentration in mol/L = \(\frac{\text{mass in g/L}}{\text{molar mass}}\) = \(\frac{1}{100}\) mol/L = 10^-2 mol/L

Thus, the concentration of the solution in mol/L is 10^-2 mol/L